Question

随机变量x服从几何分布,其分布律为P(x=k)=p(1-p)^(k-1),k=1,2...,求E(x),D(x)，求详解步骤

Answer 1

下面的计算利用幂级数展开式(通过1/(1-x)=∑{k,0,∞}x^k，x∈(-1,1)容易证明) ：
1/(1-x)²=1+2x+3x²+4x³+…=∑{k,0,∞}(k+1)*x^k，x∈(-1,1) ①

注意到0<1-p<1
E(X)=∑{k,1,∞}k*p*(1-p)^(k-1)
=p*∑{k,0,∞}(k+1)*(1-p)^k
=p*1/[1-(1-p)]² 由①
=1/p

为计算D(X)，可先求出幂级数∑{k,0,∞}(k+1)²*x^k，x∈(-1,1)的和函数
令S(x)=∑{k,0,∞}(k+1)²*x^k，x∈(-1,1)，则
∫{0,x}S(x)dx=∫{0,x}[∑{k,0,∞}(k+1)²*x^k]dx
=∑{k,0,∞}∫{0,x} [(k+1)²*x^k]dx
=∑{k,0,∞}(k+1)*x^(k+1)
=x*∑{k,0,∞}(k+1)*x^k
=x*1/(1-x)² 由①
S(x)=[ x/(1-x)²]'=(1+x)/(1-x)³，即
∑{k,0,∞}(k+1)²*x^k=(1+x)/(1-x)³，x∈(-1,1) ②

∵E(X²)=∑{k,1,∞}k²*p*(1-p)^(k-1)
=p*∑{k,0,∞}(k+1)²*(1-p)^k
=p*[1+(1-p)]/[1-(1-p)]³ 由②
=(2-p)/p²

∴D(X)= E(X²)-[E(X)]²
=(2-p)/p²-1/p²
=(1-p)/p²